2004 AMC 10B Problem 25

The shaded region is awkward because of the hole. So fill the hole first: find the whole overlap of the two big circles, then subtract the small circle (area $\pi \cdot 1^2 = \pi$).

1. Where are the big centers? The big circle that touches the small one at $B$ has radius $2$ and its center lies on the line through the small circle's center, $2$ units from $B$. That is exactly $A$, the other end of the diameter. Likewise the other big circle is centered at $B$. So we have two circles of radius $2$ centered at $A$ and $B$, with $AB = 2$.
2. The crossing points. Let $C$ and $D$ be where the two big circles cross. $C$ is on both circles, so $CA = 2$ and $CB = 2$. With $AB = 2$, triangle $ABC$ has all sides $2$: it is equilateral. The same holds for $ABD$.
3. The angle at $A$. $\angle CAB = 60^\circ$ and $\angle BAD = 60^\circ$, so $\angle CAD = 120^\circ$.
4. Split the overlap. The straight line $CD$ cuts the football-shaped overlap into two equal halves. The half nearer $B$ is part of the circle centered at $A$: it is the $120^\circ$ slice $ACD$ with triangle $ACD$ removed.
- Slice: $120^\circ$ is $\frac{1}{3}$ of the circle, so its area is $\frac{1}{3} \cdot \pi \cdot 2^2 = \frac{4\pi}{3}$.
- Triangle $ACD$: the line $CD$ crosses $AB$ at its middle, $1$ from $A$, so the height from $A$ is $1$. The length $CD$ is twice the height of an equilateral triangle of side $2$, which is $2\sqrt{3}$. Area $= \frac{1}{2} \cdot 2\sqrt{3} \cdot 1 = \sqrt{3}$.
- One half of the overlap: $\frac{4\pi}{3} - \sqrt{3}$.
5. Whole overlap: $2\left(\frac{4\pi}{3} - \sqrt{3}\right) = \frac{8\pi}{3} - 2\sqrt{3}$.
6. Take out the small circle:
$$\frac{8\pi}{3} - 2\sqrt{3} - \pi = \frac{5\pi}{3} - 2\sqrt{3}.$$

The answer is $\boxed{\textbf{(B)}\ \frac{5}{3}\pi - 2\sqrt{3}}$.

A region between arcs is built from pieces of the form "slice of a circle minus a triangle." Drawing lines from the crossing points to the centers shows the angles; equal radii and center distance make equilateral triangles, so the angles are $60^\circ$ and $120^\circ$ and $\sqrt{3}$ shows up. When a region has a hole, find the region without the hole first and subtract the hole at the end.

If time is short, use the choices. Choices (C), (D), (E) are each exactly $\pi$ more than (A) or (B), so they match forgetting to remove the small circle. The angles here are $60^\circ$ and $120^\circ$, which bring $\sqrt{3}$, not $\sqrt{2}$. That leaves (B).